Problem 4

Exercise 6.3 (Indicator function limit) Let $f : [0,1]\to\mathbb{R}$ be continuous and $g:[0,1]\to[0,1]$ measurable. Compute the limit

$\displaystyle \lim_{n\to\infty} \int_0^1 f(g(x)^n) dx$

(6.21)

Proof. The function $f$

has domain a compact set, so $f$

is bounded. The domain of integration has finite measure and the function $f\circ g^n$ is measurable, so the bounded convergence theorem allows the limit interchange

$\displaystyle \lim_{n\to\infty} \int_0^1 f(g(x)^n) dx = \int_0^1 \lim_{n\to\infty} f(g(x)^n) dx$

(6.22)

As $n\to\infty$ the function $g^n$

becomes an indicator function

$\displaystyle \lim_{n\to\infty} g(x)^n = \begin{cases} 1 \quad x\in E\\ 0 \quad x\in[0,1]\setminus E \end{cases}$

(6.23)

where $E = g^{-1}(\{1\})$ . The continuity of $f$

then reveals $f\circ g^n$ becomes an indicator function

$\displaystyle \lim_{n\to\infty} f(g(x)^n) = \begin{cases} f(1) \quad x\in E\\ f(0) \quad x\in[0,1]\setminus E \end{cases}$

(6.24)

Therefore,

$\displaystyle \lim_{n\to\infty} \int_0^1 f(g(x)^n)$	$\displaystyle = \int_0^1 \begin{cases} f(1) \quad x\in E\\ f(0) \quad x\in[0,1]\setminus E \end{cases}dx$	(6.25)
	$\displaystyle = \int_E f(1)dx + \int_{[0,1]\setminus E} f(0)dx$	(6.26)
	$\displaystyle = \mu(E)f(1) + \mu([0,1]\setminus E)f(0)$	(6.27)

$\qedsymbol$